A catapult on a cliff launches a large round rock towards a ship on the ocean be
ID: 1476468 • Letter: A
Question
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H of 32.0 m above sea level, directed at an angle above the horizontal with an unknown speed v0.The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 166.0 m. Assuming that air friction can be neglected, calculate the value of the angle .Calculate the speed at which the rock is launched.To what height above sea level does the rock rise?
Please ANSWER all three parts
Explanation / Answer
from the goven data,
vox = 166/6
= 27.7 m/s
Apply, -h = voy*t - 0.5*g*t^2
-32 = voy*6 - 0.5*9.8*6^2
voy = (0.5*9.8*6^2 - 32)/6
= 24 m/s
theta = tan^-1(voy/vox)
= tan^-1(24/27.7)
= 41 degrees <<<<<<<<<-------------------Answer
vo = sqrt(vox^2 + voy^2)
= sqrt(27.7^2 + 24^2)
= 36.7 m/s <<<<<<<<<-------------------Answer
Hmax = h + voy^2/(2*g)
= 32 + 24^2/(2*9.8)
= 61.4 m <<<<<<<<<-------------------Answer
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